LeetCode #2035 — HARD

Partition Array Into Two Arrays to Minimize Sum Difference

Break down a hard problem into reliable checkpoints, edge-case handling, and complexity trade-offs.

Solve on LeetCode

The Problem

Problem Statement

You are given an integer array nums of 2 * n integers. You need to partition nums into two arrays of length n to minimize the absolute difference of the sums of the arrays. To partition nums, put each element of nums into one of the two arrays.

Return the minimum possible absolute difference.

Example 1:

Input: nums = [3,9,7,3]
Output: 2
Explanation: One optimal partition is: [3,9] and [7,3].
The absolute difference between the sums of the arrays is abs((3 + 9) - (7 + 3)) = 2.

Example 2:

Input: nums = [-36,36]
Output: 72
Explanation: One optimal partition is: [-36] and [36].
The absolute difference between the sums of the arrays is abs((-36) - (36)) = 72.

Example 3:

Input: nums = [2,-1,0,4,-2,-9]
Output: 0
Explanation: One optimal partition is: [2,4,-9] and [-1,0,-2].
The absolute difference between the sums of the arrays is abs((2 + 4 + -9) - (-1 + 0 + -2)) = 0.

Constraints:

1 <= n <= 15
nums.length == 2 * n
-10⁷ <= nums[i] <= 10⁷

Roadmap

Brute Force Baseline
Core Insight
Algorithm Walkthrough
Edge Cases
Full Annotated Code
Interactive Study Demo
Complexity Analysis

Step 01

Brute Force Baseline

Problem summary: You are given an integer array nums of 2 * n integers. You need to partition nums into two arrays of length n to minimize the absolute difference of the sums of the arrays. To partition nums, put each element of nums into one of the two arrays. Return the minimum possible absolute difference.

Baseline thinking

Start with the most direct exhaustive search. That gives a correctness anchor before optimizing.

Pattern signal: Array · Two Pointers · Binary Search · Dynamic Programming · Bit Manipulation · Segment Tree

Example 1

[3,9,7,3]

Example 2

[-36,36]

Example 3

[2,-1,0,4,-2,-9]

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Last Stone Weight II (last-stone-weight-ii)
Fair Distribution of Cookies (fair-distribution-of-cookies)

Step 02

Core Insight

What unlocks the optimal approach

The target sum for the two partitions is sum(nums) / 2.
Could you reduce the time complexity if you arbitrarily divide nums into two halves (two arrays)? Meet-in-the-Middle?
For both halves, pre-calculate a 2D array where the kth index will store all possible sum values if only k elements from this half are added.
For each sum of k elements in the first half, find the best sum of n-k elements in the second half such that the two sums add up to a value closest to the target sum from hint 1. These two subsets will form one array of the partition.

Interview move: turn each hint into an invariant you can check after every iteration/recursion step.

Step 03

Algorithm Walkthrough

Iteration Checklist

Define state (indices, window, stack, map, DP cell, or recursion frame).
Apply one transition step and update the invariant.
Record answer candidate when condition is met.
Continue until all input is consumed.

Use the first example testcase as your mental trace to verify each transition.

Step 04

Edge Cases

Minimum Input

Single element / shortest valid input

Validate boundary behavior before entering the main loop or recursion.

Duplicates & Repeats

Repeated values / repeated states

Decide whether duplicates should be merged, skipped, or counted explicitly.

Extreme Constraints

Largest constraint values

Re-check complexity target against constraints to avoid time-limit issues.

Invalid / Corner Shape

Empty collections, zeros, or disconnected structures

Handle special-case structure before the core algorithm path.

Step 05

Full Annotated Code

Source-backed implementations are provided below for direct study and interview prep.

// Accepted solution for LeetCode #2035: Partition Array Into Two Arrays to Minimize Sum Difference
class Solution {
    public int minimumDifference(int[] nums) {
        int n = nums.length >> 1;
        Map<Integer, Set<Integer>> f = new HashMap<>();
        Map<Integer, Set<Integer>> g = new HashMap<>();
        for (int i = 0; i < (1 << n); ++i) {
            int s = 0, cnt = 0;
            int s1 = 0, cnt1 = 0;
            for (int j = 0; j < n; ++j) {
                if ((i & (1 << j)) != 0) {
                    s += nums[j];
                    ++cnt;
                    s1 += nums[n + j];
                    ++cnt1;
                } else {
                    s -= nums[j];
                    s1 -= nums[n + j];
                }
            }
            f.computeIfAbsent(cnt, k -> new HashSet<>()).add(s);
            g.computeIfAbsent(cnt1, k -> new HashSet<>()).add(s1);
        }
        int ans = Integer.MAX_VALUE;
        for (int i = 0; i <= n; ++i) {
            List<Integer> fi = new ArrayList<>(f.get(i));
            List<Integer> gi = new ArrayList<>(g.get(n - i));
            Collections.sort(fi);
            Collections.sort(gi);
            for (int a : fi) {
                int left = 0, right = gi.size() - 1;
                int b = -a;
                while (left < right) {
                    int mid = (left + right) >> 1;
                    if (gi.get(mid) >= b) {
                        right = mid;
                    } else {
                        left = mid + 1;
                    }
                }
                ans = Math.min(ans, Math.abs(a + gi.get(left)));
                if (left > 0) {
                    ans = Math.min(ans, Math.abs(a + gi.get(left - 1)));
                }
            }
        }
        return ans;
    }
}

// Accepted solution for LeetCode #2035: Partition Array Into Two Arrays to Minimize Sum Difference
func minimumDifference(nums []int) int {
	n := len(nums) >> 1
	f := make([][]int, n+1)
	g := make([][]int, n+1)
	for i := 0; i < (1 << n); i++ {
		s, cnt := 0, 0
		s1, cnt1 := 0, 0
		for j := 0; j < n; j++ {
			if (i & (1 << j)) != 0 {
				s += nums[j]
				cnt++
				s1 += nums[n+j]
				cnt1++
			} else {
				s -= nums[j]
				s1 -= nums[n+j]
			}
		}
		f[cnt] = append(f[cnt], s)
		g[cnt1] = append(g[cnt1], s1)
	}

	for i := 0; i <= n; i++ {
		sort.Ints(f[i])
		sort.Ints(g[i])
	}
	ans := 1 << 30
	for i := 0; i <= n; i++ {
		for _, a := range f[i] {
			left, right := 0, len(g[n-i])-1
			b := -a
			for left < right {
				mid := (left + right) >> 1
				if g[n-i][mid] >= b {
					right = mid
				} else {
					left = mid + 1
				}
			}
			ans = min(ans, abs(a+g[n-i][left]))
			if left > 0 {
				ans = min(ans, abs(a+g[n-i][left-1]))
			}
		}
	}
	return ans
}

func abs(x int) int {
	if x > 0 {
		return x
	}
	return -x
}

# Accepted solution for LeetCode #2035: Partition Array Into Two Arrays to Minimize Sum Difference
class Solution:
    def minimumDifference(self, nums: List[int]) -> int:
        n = len(nums) >> 1
        f = defaultdict(set)
        g = defaultdict(set)
        for i in range(1 << n):
            s = cnt = 0
            s1 = cnt1 = 0
            for j in range(n):
                if (i & (1 << j)) != 0:
                    s += nums[j]
                    cnt += 1
                    s1 += nums[n + j]
                    cnt1 += 1
                else:
                    s -= nums[j]
                    s1 -= nums[n + j]
            f[cnt].add(s)
            g[cnt1].add(s1)

        ans = inf
        for i in range(n + 1):
            fi, gi = sorted(list(f[i])), sorted(list(g[n - i]))
            # min(abs(f[i] + g[n - i]))
            for a in fi:
                left, right = 0, len(gi) - 1
                b = -a
                while left < right:
                    mid = (left + right) >> 1
                    if gi[mid] >= b:
                        right = mid
                    else:
                        left = mid + 1
                ans = min(ans, abs(a + gi[left]))
                if left > 0:
                    ans = min(ans, abs(a + gi[left - 1]))
        return ans

// Accepted solution for LeetCode #2035: Partition Array Into Two Arrays to Minimize Sum Difference
/**
 * [2035] Partition Array Into Two Arrays to Minimize Sum Difference
 *
 * You are given an integer array nums of 2 * n integers. You need to partition nums into two arrays of length n to minimize the absolute difference of the sums of the arrays. To partition nums, put each element of nums into one of the two arrays.
 * Return the minimum possible absolute difference.
 *  
 * Example 1:
 * <img alt="example-1" src="https://assets.leetcode.com/uploads/2021/10/02/ex1.png" style="width: 240px; height: 106px;" />
 * Input: nums = [3,9,7,3]
 * Output: 2
 * Explanation: One optimal partition is: [3,9] and [7,3].
 * The absolute difference between the sums of the arrays is abs((3 + 9) - (7 + 3)) = 2.
 *
 * Example 2:
 *
 * Input: nums = [-36,36]
 * Output: 72
 * Explanation: One optimal partition is: [-36] and [36].
 * The absolute difference between the sums of the arrays is abs((-36) - (36)) = 72.
 *
 * Example 3:
 * <img alt="example-3" src="https://assets.leetcode.com/uploads/2021/10/02/ex3.png" style="width: 316px; height: 106px;" />
 * Input: nums = [2,-1,0,4,-2,-9]
 * Output: 0
 * Explanation: One optimal partition is: [2,4,-9] and [-1,0,-2].
 * The absolute difference between the sums of the arrays is abs((2 + 4 + -9) - (-1 + 0 + -2)) = 0.
 *
 *  
 * Constraints:
 *
 * 	1 <= n <= 15
 * 	nums.length == 2 * n
 * 	-10^7 <= nums[i] <= 10^7
 *
 */
pub struct Solution {}

// problem: https://leetcode.com/problems/partition-array-into-two-arrays-to-minimize-sum-difference/
// discuss: https://leetcode.com/problems/partition-array-into-two-arrays-to-minimize-sum-difference/discuss/?currentPage=1&orderBy=most_votes&query=

// submission codes start here

impl Solution {
    pub fn minimum_difference(nums: Vec<i32>) -> i32 {
        0
    }
}

// submission codes end

#[cfg(test)]
mod tests {
    use super::*;

    #[test]
    #[ignore]
    fn test_2035_example_1() {
        let nums = vec![3, 9, 7, 3];

        let result = 2;

        assert_eq!(Solution::minimum_difference(nums), result);
    }

    #[test]
    #[ignore]
    fn test_2035_example_2() {
        let nums = vec![-36, 36];

        let result = 72;

        assert_eq!(Solution::minimum_difference(nums), result);
    }

    #[test]
    #[ignore]
    fn test_2035_example_3() {
        let nums = vec![2, -1, 0, 4, -2, -9];

        let result = 0;

        assert_eq!(Solution::minimum_difference(nums), result);
    }
}

// Accepted solution for LeetCode #2035: Partition Array Into Two Arrays to Minimize Sum Difference
// Auto-generated TypeScript example from java.
function exampleSolution(): void {
}
// Reference (java):
// // Accepted solution for LeetCode #2035: Partition Array Into Two Arrays to Minimize Sum Difference
// class Solution {
//     public int minimumDifference(int[] nums) {
//         int n = nums.length >> 1;
//         Map<Integer, Set<Integer>> f = new HashMap<>();
//         Map<Integer, Set<Integer>> g = new HashMap<>();
//         for (int i = 0; i < (1 << n); ++i) {
//             int s = 0, cnt = 0;
//             int s1 = 0, cnt1 = 0;
//             for (int j = 0; j < n; ++j) {
//                 if ((i & (1 << j)) != 0) {
//                     s += nums[j];
//                     ++cnt;
//                     s1 += nums[n + j];
//                     ++cnt1;
//                 } else {
//                     s -= nums[j];
//                     s1 -= nums[n + j];
//                 }
//             }
//             f.computeIfAbsent(cnt, k -> new HashSet<>()).add(s);
//             g.computeIfAbsent(cnt1, k -> new HashSet<>()).add(s1);
//         }
//         int ans = Integer.MAX_VALUE;
//         for (int i = 0; i <= n; ++i) {
//             List<Integer> fi = new ArrayList<>(f.get(i));
//             List<Integer> gi = new ArrayList<>(g.get(n - i));
//             Collections.sort(fi);
//             Collections.sort(gi);
//             for (int a : fi) {
//                 int left = 0, right = gi.size() - 1;
//                 int b = -a;
//                 while (left < right) {
//                     int mid = (left + right) >> 1;
//                     if (gi.get(mid) >= b) {
//                         right = mid;
//                     } else {
//                         left = mid + 1;
//                     }
//                 }
//                 ans = Math.min(ans, Math.abs(a + gi.get(left)));
//                 if (left > 0) {
//                     ans = Math.min(ans, Math.abs(a + gi.get(left - 1)));
//                 }
//             }
//         }
//         return ans;
//     }
// }

Step 06

Interactive Study Demo

Use this to step through a reusable interview workflow for this problem.

Custom checkpoints (one per line)

Press Step or Run All to begin.

Step 07

Complexity Analysis

Time

O(n)

Space

O(1)

Approach Breakdown

BRUTE FORCE

O(n²) time

O(1) space

Two nested loops check every pair of elements. The outer loop picks one element, the inner loop scans the rest. For n elements that is n × (n−1)/2 comparisons = O(n²). No extra memory — just two loop variables.

TWO POINTERS

O(n) time

O(1) space

Each pointer traverses the array at most once. With two pointers moving inward (or both moving right), the total number of steps is bounded by n. Each comparison is O(1), giving O(n) overall. No auxiliary data structures are needed — just two index variables.

Shortcut: Two converging pointers on sorted data → O(n) time, O(1) space.

Coach Notes

Common Mistakes

Review these before coding to avoid predictable interview regressions.

Off-by-one on range boundaries

Wrong move: Loop endpoints miss first/last candidate.

Usually fails on: Fails on minimal arrays and exact-boundary answers.

Fix: Re-derive loops from inclusive/exclusive ranges before coding.

Moving both pointers on every comparison

Wrong move: Advancing both pointers shrinks the search space too aggressively and skips candidates.

Usually fails on: A valid pair can be skipped when only one side should move.

Fix: Move exactly one pointer per decision branch based on invariant.

Boundary update without `+1` / `-1`

Wrong move: Setting `lo = mid` or `hi = mid` can stall and create an infinite loop.

Usually fails on: Two-element ranges never converge.

Fix: Use `lo = mid + 1` or `hi = mid - 1` where appropriate.

State misses one required dimension

Wrong move: An incomplete state merges distinct subproblems and caches incorrect answers.

Usually fails on: Correctness breaks on cases that differ only in hidden state.

Fix: Define state so each unique subproblem maps to one DP cell.